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In S. Boyd textbook p.105 (button): (cvx = convex)

Let F = log f & G = log g are convex (i.e. Let f & g are log-cvx)

(This guarantees f & g are cvx, since log-cvx is included in cvx)

Now, the book says:

From the composition rules for cvx functions: log(exp(F) + exp(G)) = log(f + g) is cvx.

I do not understand the last step?

What I think is:

exp(F) & exp(G) are cvx .... (OK)

exp(F) + exp(G) are cvx .... (OK)

However, logarithm is a concave function in R++

Composition rules:

enter image description here

Thanks!!

sleeve chen
  • 8,281

2 Answers2

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I was puzzled by the statement too. Here is my proof:

$F$ and $G$ are convex. With $e(*) = \exp(*)$, $$Z=\log[e(F) + e(G)]$$ $$Z' = \frac{[e(F)F' + e(G)G']}{[e(F) + e(G)]}$$ $$Z" = -\frac{[e(F)F' + e(G)G']^2}{[e(F) + e(G)]^2}+\frac{[e(F)(F')^2+ e(F)F''+ e(G)(G')^2+e(G)G'']}{[e(F) + e(G)]}$$ Check: $$T = [e(F)(F')^2+ e(F)F''+ e(G)(G')²+e(G)G'']\cdot [e(F) + e(G)] - [e(F)F' + e(G)G']^2$$ $$T = e(F)(F')^2(G) + e(G)(G')^2e(F) - 2e(G)e(F)F'G' + [e(F)F''+e(G)G'']\cdot [e(F)+e(G)]$$ $$= e(F)e(G)[F'-G']^2+[e(F)F''+e(G)G'']\cdot [e(F)+e(G)]$$ $$e(f)>0, \ e(G)>0, \ F''\ge 0, \ G''\ge 0.$$
So: $$T\ge0 \implies Z''\ge0 \implies Z \ \text{is convex}$$ Also: $$Z=\log(f+g)$$ $$\implies f+g \ \text{is log-convex.}$$

user26857
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jye
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There's a nice proof using a baby version of Hölder’s inequality.

The log-convexity condition on a positive function $f$ is equivalent to the inequality

$$f\left(\frac1{p}x + \frac1{q}y\right) \leq f(x)^{\frac1{p}} \cdot f(y)^\frac1{q}$$ for all $x, y \in \mathrm{dom}(f)$ and all positive numbers $p, q$ satisfying $\frac1{p} + \frac1{q} = 1$. The version of Hölder’s inequality we'll use states that for positive numbers $s, t, u, v$,

$$su + tv \leq (s^p + t^p)^\frac1{p} \cdot (u^q + v^q)^\frac1{q}$$ (which is a special case of the usual Hölder’s inequality by considering a two-point measure space with counting measure -- clearly overkill, but the connection with Hölder is irresistible to point out).

For log-convex functions $f, g$ and points $x, y$ in their common domain, put

$$s = f(x)^\frac1{p}, \qquad t = g(x)^\frac1{p}, \qquad u = f(y)^\frac1{q}, \qquad v = g(y)^\frac1{q}.$$ Then log-convexity of $f + g$ follows from

$$\begin{array}{ccc} f\left(\frac1{p}x + \frac1{q}y\right) + g\left(\frac1{p}x + \frac1{q}y\right) & \leq & f(x)^\frac1{p} f(y)^\frac1{q} + g(x)^\frac1{p} g(y)^\frac1{q} \\ & \leq & (f(x) + g(x))^\frac1{p} \cdot (f(y) + g(y))^\frac1{q} \end{array}$$ where the first line used log-convexity of $f$ and $g$, and the second is by Hölder. Done!

user43208
  • 8,289